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3 years ago

the atomic number of barium is 56 what do you know about the subatomic particles in an atom of this element

Chemistry

2 answers:

Yuliya22 [10]3 years ago

5 0

If this atom is a neutral atom, the subatomic particles would be 56 protons, 56 neutrons, and 56 electrons.

Reptile [31]3 years ago

4 0

Answer is 56 protons and 56 electrons.

Explanation;

Atomic number is equal to number of protons. Hence, when the atomic number is 56, it means that atom has 56 protons.

When the element is in neutral state, number of protons = number of electrons. Hence, we can say that barium atom has 56 electrons.

But same element can have different number of neutrons. Those are called isotopes. Hence, we cannot say that there are 56 neutrons in barium atom without having its mass number. (Mass number = number of protons + number of neutrons)

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Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.

insens350 [35]

Given:
Iron, 125 grams T
1 = 23.5 degrees Celsius, T2 = 78 degrees Celsius.

Required:
Heat produced in kilojoules

Solution:
The molar mass of iron is 55.8 grams per mole. SO we need to change the given mass of iron into moles.

Number of moles of iron = 125 g/(55.8 g/mol) = 2.24 moles

Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2.24 moles (8.314 Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
Q (heat) = 1014.97 Joules or 1.015 kilojoules This is the amount of heat produced in warming 125 g f iron.

7 0

3 years ago

Determine the mass fraction of iron in its compounds:

Shkiper50 [21]

Explanation:

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m=3.26 g=0.00326 kg" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; font-family: inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">m=3.26 g=0.00326 kgm=3.26 g=0.00326 kg

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IT'S TOTAL ANSWER OF ITS AND THIS QUESTION IS IN MATHEMATION FINAL EXAM. PLEASE GIVE❤ AND MARK ME A BRAINLIST

7 0

2 years ago

Rubbing alcohol evaporates from your hand quickly, leaving a cooling effect on your skin. Because evaporation is an example of a

Ludmilka [50]

The molecules in gas are farther apart and have more room to bounce around than liquid

3 0

3 years ago

Read 2 more answers

Which feature forms at convergent

geniusboy [140]

B trench

Explanation:

The feature found at most convergent margins is a trench.

A trench is a large depression that typifies most convergent margins.

Oceanic trenches are natural topographic depressions found on the sea floor.
These depressions forms where two plate boundaries converges.
The denser one slides beneath the less dense one into the asthenosphere below.
This is called a subduction zone.
The margin between the two plates that are depression is a trench.

learn more:

Lithosphere brainly.com/question/2998243

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8 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago